Question

show how you would connect three resistors each of resistance 6 ohm so that the combination has a resistance of (i) 9 ohm(ii) 4 ohm

Answer 1

Answer:

To obtain a combined resistance of 9 ohms, three resistors each of resistance 6 ohm should be connected in a parallel circuit and one in series.  

In a parallel circuit, the equivalent resistance of all the resistors connected is equal to the reciprocal value of the sum of the individual resistances.

In a series circuit, the equivalent resistance of all the resistors connected is equal to the sum of the resistance values of each of the individual resistor.

So in a parallel circuit, total resistance = $\frac{1}{Rt}=\frac{1}{R 1}+\frac{1}{R 2}+\frac{1}{R 3}$  or  

R t= $\frac{\frac{1}{2}}{R 1}+\frac{1}{R 2}+\frac{1}{R 3}$.  

In a series circuit, total resistance  = R1 +R2 + R3

(i) To get 9 ohms, two resistors should be connected in parallel and one in series.

So,

Rt in parallel will be  

Rt = $\frac{1}{\frac{1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}}=\frac{1}{\frac{1}{6}+\frac{1}{6}}$

= 3 ohms

One resistor is connected in series with the 3 ohms, so  

Rt in series will be Rt = R1 +R2 = 6+3 = 9 ohms

(ii) To obtain a combined resistance of 4 ohms, two resistors (of 6 ohms each) are connected in series and one in parallel to the series.

So,

Rt in series = R1 +R2 = 6+6 = 12 ohms

Rt in parallel = $\frac{1}{R t}=\frac{1}{R 1}+\frac{1}{R 2}=\frac{1}{6}+\frac{1}{12}$= 4 Ohms

Answer 2

Given:

A set of three resistors each of 6 Ohms.

To Find:

The possible combinations such that,

• The combination resistance is  9 ohm
• The combination resistance is 4 ohm

Solution:

1. We are given three resistances each of 6 ohms.

2. Consider the first case,

(i) 9 ohm

• Consider the three resistances as A, B, and C.
• Arrange resistances A and B in parallel and C in series.

=> Net effective resistance = ( A, B parallel ) + ( C series ),

=> For two resistances parallel to each other, the net effective resistance is,

=>$\frac{1}{R_{eff} } =\frac{1}{R_{1} } +\frac{1}{R_{2} }$,

=> R(effective) = 6 / 2,

=> R = 3 ohms.

=>Series resistance of 6ohms and 3 ohms is,

=> R(effective) = R1 + R2,

=> R (eff) = 9 ohms.

3. Consider the second case,

(ii) 4 ohms

=> Arrange the resistances A and B in series, Let C be parallel to A and B.

=> Effective resistance of A and B = 6 + 6 = 12 ohms.

=> Total effective resistance = 12ohms parallel to 6 ohms,

=>$\frac{1}{R_{eff} } =\frac{1}{R_{1} } +\frac{1}{R_{2} }$,

=> R(effective) = 12/3 = 4 ohms.

Hence, both the cases are shown.